Hi, I'm looking for some help with complex numbers. I've attached a question below with the solution.
I understand where the route 41 is coming from but I am getting a different angle.
Part (b) I am also confused by.
Help is much appreciated.
Hi, I'm looking for some help with complex numbers. I've attached a question below with the solution.
I understand where the route 41 is coming from but I am getting a different angle.
Part (b) I am also confused by.
Help is much appreciated.
Do you know that sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)?
5cos(x)- 4 sin(x) cannot be "cos(a+ b)" for any a and b because "5" and "4" cannot be "sin(a)" and "cos(a)" for any a. sin(a) and cos(a) must be between -1 and 1 for all a and $\displaystyle sin^2(a)+ cos^2(a)= 1$. However, I note that $\displaystyle 5^2+ 4^2= 25+ 16= 41$ so that
$\displaystyle 5 cos(x)- 4 sin(x)= \sqrt{41}\left(\frac{5}{\sqrt{41}}cos(x)- \frac{4}{\sqrt{41}}sin(x)\right)$.
Now, what angle, a, has $\displaystyle sin(a)= \frac{5}{\sqrt{41}}$ and $\displaystyle cos(a)= -\frac{4}{\sqrt{41}}$?
No, this is the same angle a for both. Sine is positive in the first and second quadrants while cosine is negative in the second and third quadrants. In order that sine be positive and cosine negative, the angle must be in the second quadrant- 180- 51.3= 128.7 degrees.
$\displaystyle sin(128.7)= 0.780= \frac{5}{\sqrt{41}}$ and $\displaystyle cos(128.7)= -0.63= -\frac{4}{\sqrt{41}}$
Now, remember the point was that $\displaystyle 5cos(x)- 4sin(x)= \sqrt{41}(sin(a) cos(x)+ cos(a)sin(x))= \sqrt{41}sin(x+ a)$